JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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The value of

\begin{array}{l} \left({ }^{21} \mathrm{C}_{1}-{ }^{10} \mathrm{C}_{1}\right)+\left({ }^{21} \mathrm{C}_{2}-{ }^{10} \mathrm{C}_{2}\right)+\left({ }^{21} \mathrm{C}_{3}-{ }^{10} \mathrm{C}_{3}\right)+\left({ }^{21} \mathrm{C}_{4}-{ }^{10} \mathrm{C}_{4}\right) \\ +\ldots .+\left({ }^{21} \mathrm{C}_{10}-{ }^{10} \mathrm{C}_{10}\right) \text { is : } \end{array}

View answer and explanation

Correct option: A

2^{20}-2^{10}

Option A (

2^{20}-2^{10}

) is correct because it involves summing binomial coefficients and using properties of binomial expansions.

More Questions from BINOMIAL THEOREM

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\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

a

d

\left( {n + 1} \right)

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

(1-x)^{201}\left(1+x+x^2\right)^{200}

\left(\frac{1}{x}+2 x\right)^n

{x^5}

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170

n